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Description
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
An alternative formula for the Fibonacci sequence is
.
Given an integer n, your goal is to compute the last 4 digits of Fn.
Input
The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.
Output
For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
Sample Input
0 9 999999999 1000000000 -1
Sample Output
0 34 626 6875
矩阵快速幂入门题，构造矩阵[fn,fn-1]*[1,1]在自己敲一遍模板就行了。
[0 ,0 ] [1,0]
#include<cstdio> #include<iostream> #include<algorithm> #include<cstring> typedef long long int LL; using namespace std; int n,mod=10000; struct Matrix { int a[2][2]; Matrix(){memset(a,0,sizeof(a));} Matrix operator* (const Matrix &p) { Matrix res; for(int i=0;i<2;i++) { for(int j=0;j<2;j++) { for(int k=0;k<2;k++) { res.a[i][j]+=(a[i][k]*p.a[k][j]%mod); } res.a[i][j]%=mod; } } return res; } }ans,base; Matrix quick_pow(Matrix base,int k) { Matrix res; for(int i=0;i<2;i++) { res.a[i][i]=1; } while(k) { if(k&1) res=res*base; base=base*base; k>>=1; } return res; } void init_Matrix() { ans.a[0][0]=1; ans.a[0][1]=0; ans.a[1][0]=0; ans.a[1][1]=0; base.a[0][0]=1; base.a[0][1]=1; base.a[1][0]=1; base.a[1][1]=0; } int main() { while(scanf("%d",&n)&&n!=-1) { init_Matrix(); if(n==0) printf("0\n"); else if(n==1) printf("1\n"); else { ans=ans*quick_pow(base,n-1); printf("%d\n",ans.a[0][0]); } } return 0; }
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原文地址：http://www.cnblogs.com/PrimeXuan/p/5521096.html
